The sum of a two-digit number and the number obtained by reversing the digits is always divisible by
11
Consider the number ab
ab = 10×a + b
Reverse of ab is ba.
ba = 10×b + a
Sum of the number and its reverse is,
ab + ba = 10×a + b + 10×b + a
⇒ 10×(a + b) + (a + b) = 11× (a + b)
It is always divisible by 11.